1. What is the specific heat of a 10. g sample of a substance that requires 950 J of heat to raise its temp from 18 C to 42 C? 2. Hg has a specific heat of 0.14 J/gC. How much energy does a 22.8 g sample of mercury lose as its temp goes from 33 C to 16 C ?

V. Calorimetry Calorimeter – an instrument that measures the gain or loss of heat. (See pg. 66)

A calorimeter can be used to determine the specific heat of a substance, or to determine the heat given off or absorbed by a chemical rxn. In calorimetry, the heat change for a process is determined by measuring the heat change in the water surrounding the process in an insulated container called a calorimeter.

According to the law of conservation of energy: the heat lost by the process is gained by the water (or the heat absorbed by the process is lost by the water.) In equation form this looks like: qprocess = -qwater (The specific heat of water is a constant: 4.184 J/gC)

Determining the specific heat of an object using calorimetry: heat lost by object = - heat gained by water m x cp x T = - m x cp x T (object) (water) Solve for cp of the object. *The right side of a calorimetry problem always starts with a negative sign.

EX1: A 28.5 g sample of metal with a temperature of 100.0 C is placed in a calorimeter holding 150.0 g of water at 0.00 C. The temperature of the water rises to 2.00 C. What is the specific heat of the metal? What metal is it? (use appendix A; Table A-3 in back of text) (For these problems, the final temperature of the water and the object are the same.) (Remember that the specific heat of water is always 4.184 J/gºC)

EX2: A 12.5 g sample of metal with a temperature of 100.0 C is placed in a calorimeter holding 75.0 g of water at 22.0 C. The temperature of the water rises to 24.0 C. What is the specific heat of the metal? What metal is it?

B. Determining the heat given off or absorbed by a rxn using calorimetry: H = heat given off or absorbed by a chemical rxn H of a rxn = - heat gained or lost by the water In a calorimeter: H = - m x cp x T (water) *Signs for H: + H endothermic rxn - H exothermic rxn

EX1: A chemical rxn causes 150.g of water in a calorimeter to go from 25.0 C to 28.5 C . How much heat was given off or absorbed by the rxn? b) Was the rxn exothermic or endothermic? EX2: A chemical rxn causes 200.g of water in a calorimeter to go from 30.0 C to 28.5 C . What was H for the rxn? b) Was the rxn exothermic or endothermic?

Thermochemical Equations Heats of Reaction and Thermochemical Equations

Heat of Reaction (H) – the amount of heat given off or absorbed in a chemical rxn. (H stands for “enthalpy” – the heat content of a system at constant pressure; H = Hproducts - Hreactants ) Signs for H: + H endothermic rxn - H exothermic rxn

B. Thermochemical Equations 2 P(s) + 3 Br2(l)  2 PBr3(l) + 278 kJ Exothermic or Endothermic? Usually the equation would be written as: 2 P(s) + 3 Br2(l)  2 PBr3(l) H = -278 kJ The above are examples of thermochemical equations – equations that include the heat given off or absorbed.

Write the equation: H2(g) + I2(g) + 53 kJ  2 HI(g) using H. H2(g) + I2(g)  2 HI(g) H = +53 kJ ***The H in an equation is only for the number of moles of reactants and products indicated by the coefficients in the equation. The actual H of an individual rxn depends on how much reactant was used.*** 2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ

Example Problems: 2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ 1. If 2.00 mol of O2 react, how much heat is produced? 2. If 5.00g of H2 is burned, how much heat is released? 3. How much heat is produced in the production of 100.g of H2O? 4. How many grams of H2 are needed to produce 1.00 x 104 kJ of heat?

2 PBr3(l)  2 P(s) + 3 Br2(l) H = +278 kJ 1. If 1200 kJ of heat was absorbed by the rxn, how many grams of Br2 was produced? 2. How much heat is needed to produce 25.0g of Br2?

End College Prep

Additional Facts About Thermochemical Equations Often fractions are used as coefficients: C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l) H = -1300 kJ 2. The H of a reverse rxn has the same value but opposite sign. 2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ 2 H2O(l)  2 H2(g) + O2(g) H = +572 kJ

3. The value of H depends on what physical state the reactants and products are in: 2 H2(g) + O2(g)  2 H2O(l) H = -572 kJ 2 H2(g) + O2(g)  2 H2O(g) H = -484 kJ (The difference is due to: H2O(g)  H2O(l) H = -88 kJ)

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